Page 1 of 3 Islamabad Campus Department of Electrical Engineering Program: B.E. (Electrical) Semester – Summer 2016 EL-322 Digital Signal Processing Assignment – 1 Solution Due Date: 28/07/2016 Marks: 20 Handout Date: 21/07/2016 Question # 1: Solve the linear system by Gauss-Jordan elimination: − + 2 − = −1 2 + − 2 − 2 = −2 − + 2 − 4 + = 1 3 − 3 = −3 Solution: The augmented matrix is: 1 −1 2 −1 −1 2 1 −2 −2 −2 −1 2 −4 1 1 3 0 0 −3 −3 -2R1+R2, 1R1+R3, -3R1+R4 1 −1 2 −1 −1 0 3 −6 0 0 0 1 −2 0 0 0 3 −6 0 0 1/3 R2, then -1 new R2+R3, -3 new R2+R4 1 −1 2 −1 −1 0 1 −2 0 0 0 0 0 0 0 0 0 0 0 0 R2+R1 1 0 0 −1 −1 0 1 −2 0 0 0 0 0 0 0 0 0 0 0 0 The corresponding system of equation is: − = −1 = − 1 − 2 = 0 = 2 Let z=s and w=t. The solution is x= t-1, y = 2s, z= s, w = t.
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Page1of3
Islamabad Campus
Department of Electrical Engineering Program: B.E. (Electrical)
Semester – Summer 2016
EL-322 Digital Signal Processing
Assignment – 1 Solution Due Date: 28/07/2016 Marks: 20 Handout Date: 21/07/2016 Question # 1: Solve the linear system by Gauss-Jordan elimination:
𝑥 − 𝑦 + 2𝑧 − 𝑤 = −1 2𝑥 + 𝑦 − 2𝑧 − 2𝑤 = −2 −𝑥 + 2𝑦 − 4𝑧 + 𝑤 = 1
3𝑥 − 3𝑤 = −3 Solution: The augmented matrix is:
1 −1 2 −1 −12 1 −2 −2 −2−1 2 −4 1 13 0 0 −3 −3
-2R1+R2, 1R1+R3, -3R1+R4 1 −1 2 −1 −10 3 −6 0 00 1 −2 0 00 3 −6 0 0
1/3 R2, then -1 new R2+R3, -3 new R2+R4 1 −1 2 −1 −10 1 −2 0 00 0 0 0 00 0 0 0 0
R2+R1 1 0 0 −1 −10 1 −2 0 00 0 0 0 00 0 0 0 0
The corresponding system of equation is: 𝑥 − 𝑤 = −1 𝑜𝑟 𝑥 = 𝑤 − 1 𝑦 − 2𝑧 = 0 𝑜𝑟 𝑦 = 2𝑧
Let z=s and w=t. The solution is x= t-1, y = 2s, z= s, w = t.
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Question # 2: Use the given information to find A:
𝐼 + 2𝐴 !! = −1 24 5
Solution:
𝐼 + 2𝐴 !! !! =1
−1.5− 2.45 −2−4 −1
= −113
5 −2−4 −1
𝐼 + 2𝐴 =−513
213
413
113
Thus:
2𝐴 =−513
213
413
113
− 1 00 1 =
−1813
213
413
−1213
Hence:
𝐴 =−913
113
213
−613
Question # 3: Use the inversion algorithm to find the inverse of the given matrix, if the inverse exists:
−1 3 −42 4 1−4 2 −9
Solution: −1 3 −42 4 1−4 2 −9
1 0 00 1 00 0 1
-1R1: 1 −3 42 4 1−4 2 −9
−1 0 00 1 00 0 1
-2R1+R2 and 4R1+R3: 1 −3 40 10 −70 −10 7
−1 0 02 1 0−4 0 1
1/10 R2:
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1 −3 4
0 1 −710
0 −10 7
−1 0 015
110 0
−4 0 1
10R2+R3: 1 −3 4
0 1 −710
0 0 0
−1 0 015
110 0
−2 1 1
Since there is a row of zeros on the left side, −1 3 −42 4 1−4 2 −9
is not invertible.
Question # 4: Find all values of the unknown constant (s) in order for A to be symmetric:
1. 𝐴 = 4 −3𝑎 + 5 −1
2. 𝐴 =𝑥 − 1 𝑥! 𝑥!0 𝑥 + 2 𝑥!0 0 𝑥 − 4
Solution:
1. 𝐴 = 4 −3𝑎 + 5 −1
For A to be symmetric,
𝑎!" = 𝑎!" 𝑜𝑟 − 3 = 𝑎 + 5, 𝑠𝑜 𝑎 = −8.
2. 𝐴 =𝑥 − 1 𝑥! 𝑥!0 𝑥 + 2 𝑥!0 0 𝑥 − 4
For A to be symmetric,
𝑎!" = 𝑎!" 𝑜𝑟 𝑎!" = 𝑎!" 𝑡ℎ𝑒𝑛, 𝑥! = 0, 𝑠𝑜 𝑥 = 0.
Good Luck
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